TITLE: Is the closed form for $\sum_{k=1}^\infty\frac{\overline{H}_k}{k^m}$ known in the literature?
QUESTION [5 upvotes]: I managed to find 

$$\sum_{k=1}^\infty\frac{\overline{H}_k}{k^m}=(1-2^{-m})\sum_{k=1}^\infty\frac{H_k}{k^m}-2^{-m}\sum_{k=1}^\infty\frac{H_k}{(k+1/2)^m}$$
$$=(1-2^{-m})\left[\left(1+\frac m2\right)\zeta(m+1)-\frac12\sum_{i=1}^{m-2}\zeta(i+1)\zeta(m-i)\right]$$
$$+\frac{(-2)^{-m-1}}{(m-1)!}\left[2\gamma\ \psi^{(m-1)}\left(\frac12\right)-\psi^{(m)}\left(\frac12\right)+\lim_{\substack{a\to1/2}}\frac{\partial^{m-1}}{\partial a^{m-1}}\psi(a)^2\right]$$

Where $\overline{H}_k$ is the skew harmonic number, $\gamma$ is Euler–Mascheroni constant, $\zeta$ is the Riemann zeta function and $\psi^{(m)}(a)$ is the Polylogarithm function where 
$$\psi^{(m)}\left(\frac12\right)=(-1)^mm!(1-2^{m+1})\zeta(m+1)$$
My question is the closed form above known in the literature? and can we do further simplifications for the limit term to have a cleaner closed form? Also I would like to see different approaches.
Thank you

Proof
$$\sum_{k=1}^\infty\frac{\overline{H}_k}{k^m}=1+\sum_{k=2}^\infty\frac{\overline{H}_k}{k^m}=1+\sum_{k=1}^\infty\frac{\overline{H}_{2k}}{(2k)^m}+\sum_{k=1}^\infty\frac{\overline{H}_{2k+1}}{(2k+1)^m}$$
By writing $\overline{H}_{2k}=H_{2k}-H_k$ and $\overline{H}_{2k+1}=H_{2k+1}-H_k$ we have
$$\sum_{k=1}^\infty\frac{\overline{H}_{2k}}{(2k)^m}=\sum_{k=1}^\infty\frac{H_{2n}}{(2n)^m}-\sum_{n=1}^\infty\frac{H_{n}}{(2n)^m}=\frac12\sum_{k=1}^\infty\frac{(-1)^kH_{k}}{k^m}+\left(\frac12-2^{-m}\right)\sum_{k=1}^\infty\frac{H_{k}}{k^4}$$
and 
$$\sum_{k=1}^\infty\frac{\overline{H}_{2k+1}}{(2k+1)^m}=\color{blue}{\sum_{k=1}^\infty\frac{H_{2k+1}}{(2k+1)^m}}-\sum_{k=1}^\infty\frac{H_k}{(2n+1)^m}$$
$$=\color{blue}{-1+\sum_{n=0}^\infty\frac{H_{2n+1}}{(2n+1)^m}}-\sum_{k=1}^\infty\frac{H_k}{(2k+1)^m}$$
$$=\color{blue}{-1+\frac12\sum_{k=0}^\infty\frac{(-1)^kH_{k+1}}{(k+1)^m}+\frac12\sum_{k=0}^\infty\frac{H_{k+1}}{(k+1)^m}}-\sum_{n=1}^\infty\frac{H_n}{(2n+1)^m}$$
$$=\color{blue}{-1-\frac12\sum_{k=1}^\infty\frac{(-1)^kH_{k}}{k^m}+\frac12\sum_{k=1}^\infty\frac{H_{k}}{k^m}}-\sum_{k=1}^\infty\frac{H_k}{(2k+1)^m}\\$$
Combine the two sums,
$$\Longrightarrow \sum_{k=1}^\infty\frac{\overline{H}_k}{k^m}=(1-2^{-m})\sum_{k=1}^\infty\frac{H_k}{k^m}-\sum_{k=1}^\infty\frac{H_k}{(2k+1)^m}\tag1$$
The first sum is well-known

$$\sum_{k=1}^\infty\frac{H_k}{k^m}=\left(1+\frac m2\right)\zeta(m+1)-\frac12\sum_{i=1}^{m-2}\zeta(i+1)\zeta(m-i)$$

For the second sum, from here we have 
$$\int_0^1\frac{x^{n}\ln^m(x)\ln(1-x)}{1-x}\ dx=(-1)^{m-1}m!\sum_{k=1}^\infty\frac{H_k}{(k+n+1)^{m+1}}\\=\frac12\frac{\partial^m}{\partial n^m}\left(H_n^2+H_n^{(2)}\right),\quad n\in\mathbb{R}\ge-1,\quad m\in\mathbb{N}$$
Let $m+1\to m$ and $n+1=a$ we get
$$(-1)^m (m-1)!\sum_{k=1}^\infty\frac{H_k}{(k+a)^m}=\frac12\frac{\partial^{m-1}}{\partial a^{m-1}}(H_{a-1}^2+H_{a-1}^{(2)})$$
Substitute $H_{a-1}=\psi(a)+\gamma$ and $H_{a-1}^{(2)}=\zeta(2)-\psi^{(1)}(a)$
$$(-1)^m (m-1)!\sum_{k=1}^\infty\frac{H_k}{(k+a)^m}=\frac12\frac{\partial^{m-1}}{\partial a^{m-1}}((\psi(a)+\gamma)^2+\zeta(2)-\psi^{(1)}(a))$$
Because $m\ge 2$ for convergence, we can ignore the constants $\gamma$ and $\zeta(2)$ on the right side,
$$(-1)^m (m-1)!\sum_{k=1}^\infty\frac{H_k}{(k+a)^m}=\frac12\frac{\partial^{m-1}}{\partial a^{m-1}}(\psi(a)^2-\psi^{(1)}(a)+2\gamma\ \psi(a))$$
$$=\frac12\left[2\gamma\ \psi^{(m-1)}(a)-\psi^{(m)}(a)+\frac{\partial^{m-1}}{\partial a^{m-1}}\psi(a)^2\right]$$
Now take the limit to both sides and let $a\to 1/2$ we get

$$\sum_{k=1}^\infty\frac{H_k}{(k+1/2)^m}=\frac{(-1)^m}{2(m-1)!}\left[2\gamma\ \psi^{(m-1)}\left(\frac12\right)-\psi^{(m)}\left(\frac12\right)+\lim_{\substack{a\to1/2}}\frac{\partial^{m-1}}{\partial a^{m-1}}\psi(a)^2\right]$$

By combining the results of the two sums, the closed form follows.

Note
I am tagging " integration" as logarithmic integrals and harmonic series are strongly related.

REPLY [3 votes]: Following robjohn's technique we have
$$S=\sum_{j=0}^k\eta(k+2-j)\eta(j+2)=\sum_{j=0}^k\left(\sum_{m=1}^\infty\frac{(-1)^{m-1}}{m^{k+2-j}}\right)\left(\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^{j+2}}\right)$$
change the order of summations
$$=\sum_{m=1}^\infty\sum_{n=1}^\infty\sum_{j=0}^k\frac{(-1)^{m+n}}{m^{k+2-j}n^{j+2}}$$
break up the middle sum
$$=\sum_{m=1}^\infty\left(a_{n=m}+\sum_{n=1}^{m-1}+\sum_{n=m+1}^\infty\right)\sum_{j=0}^k\frac{(-1)^{m+n}}{m^{k+2-j}n^{j+2}}$$
pull out the terms for $n=m$
$$=\sum_{m=1}^\infty\sum_{j=0}^k\frac1{m^{k+4}}+\sum_{m=1}^\infty\left(\sum_{n=1}^{m-1}+\sum_{n=m+1}^\infty\right)\sum_{j=0}^k\frac{(-1)^{m+n}}{m^{k+2-j}n^{j}}$$
$$=\sum_{j=0}^k\left(\sum_{m=1}^\infty\frac1{m^{k+4}} \right)+\sum_{m=1}^\infty\left(\sum_{n=1}^{m-1}+\sum_{n=m+1}^\infty\right)\frac{(-1)^{m+n}}{m^{k+2}n^2}\left(\sum_{j=0}^k\frac{m^j}{n^j}\right)$$
$$=\sum_{j=0}^k\zeta(k+4)+\sum_{m=1}^\infty\left(\sum_{n=1}^{m-1}+\sum_{n=m+1}^\infty\right)\frac{(-1)^{m+n}}{nm^{k+2}(n-m)}-\frac{(-1)^{m+n}}{mn^{k+2}(n-m)}$$
$$=(k+1)\zeta(k+4)+\sum_{m=1}^\infty\sum_{n=1}^{m-1}\frac{(-1)^{m+n}}{nm^{k+2}(n-m)}-\frac{(-1)^{m+n}}{mn^{k+2}(n-m)}$$
$$+\sum_{m=1}^\infty\sum_{n=m+1}^{\infty}\frac{(-1)^{m+n}}{nm^{k+2}(n-m)}-\frac{(-1)^{m+n}}{mn^{k+2}(n-m)}$$
By using the general change of order summation 
$$\sum_{m=1}^\infty \sum_{n=1}^{m-1}f(m,n)=\sum_{n=1}^\infty\sum_{m=n+1}^{\infty}f(m,n),$$
the first double sum becomes
$$\sum_{m=1}^\infty\sum_{n=1}^{m-1}\frac{(-1)^{m+n}}{nm^{k+1}(n-m)}-\frac{(-1)^{m+n}}{mn^{k+2}(n-m)}=\sum_{n=1}^\infty\sum_{m=n+1}^{\infty}\frac{(-1)^{m+n}}{nm^{k+2}(n-m)}-\frac{(-1)^{m+n}}{mn^{k+2}(n-m)}$$
swab the variables $n$ and $m$
$$=\sum_{m=1}^\infty\sum_{n=m+1}^{\infty}\frac{(-1)^{n+m}}{mn^{k+2}(m-n)}-\frac{(-1)^{n+m}}{nm^{k+2}(m-n)}$$
$$=\sum_{m=1}^\infty\sum_{n=m+1}^{\infty}\frac{(-1)^{n+m}}{nm^{k+2}(n-m)}-\frac{(-1)^{n+m}}{mn^{k+2}(n-m)}$$
Thus, 
$$S=(k+1)\zeta(k+4)
+2\sum_{m=1}^\infty\sum_{n=m+1}^\infty\frac{(-1)^{n+m}}{nm^{k+2}(n-m)}-\frac{(-1)^{n+m}}{mn^{k+2}(n-m)}$$
reindex the inner sum
$$=(k+1)\zeta(k+4)
+2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{(-1)^n}{(n+m)m^{k+1}n}-\frac{(-1)^n}{m(n+m)^{k+1}n}$$
$$=(k+1)\zeta(k+4)
+2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{(-1)^n}{(n+m)m^{k+2}n}-2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{(-1)^n}{m(n+m)^{k+2}n}$$
The first sum:
$$S_1=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{(-1)^n}{(n+m)m^{k+2}n}=\sum_{m=1}^\infty\frac{1}{m^{k+3}}\left(\sum_{n=1}^\infty\frac{(-1)^nm}{n(n+m)}\right)$$
where 
$$\sum_{n=1}^\infty\frac{(-1)^nm}{n(n+m)}=\sum_{n=1}^\infty\frac{(-1)^n}{n}-\sum_{n=1}^\infty\frac{(-1)^n}{n+m}$$
$$=-\ln(2)-(-1)^n\left[\overline{H}_m-\ln(2)\right]$$
Thus,
$$S_1=\sum_{m=1}^\infty\frac{1}{m^{k+3}}\left[(-1)^m \ln(2)-\ln(2)-(-1)^m\overline{H}_m\right]$$
$$=-\ln(2)\eta(k+3)-\ln(2)\zeta(k+3)-\sum_{m=1}^\infty\frac{(-1)^m\overline{H}_m}{m^{k+3}}$$
The second sum:
$$S_2=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{(-1)^n}{m(n+m)^{k+2}n}=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{(-1)^n(n+m)}{m(n+m)^{k+3}n}$$
$$=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{(-1)^n}{m(n+m)^{k+3}}+\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{(-1)^n}{(n+m)^{k+3}n}$$
swap the variables $m$ and $n$ in the first double sum and change the order of summation in the second double sum
$$=\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{(-1)^m+(-1)^n}{n(n+m)^{k+3}}$$
reindex the inner sum
$$=\sum_{n=1}^\infty\sum_{m=n+1}^\infty\frac{(-1)^{m-n}+(-1)^n}{nm^{k+3}}$$
use $\sum_{m=n+1}^\infty f(m)=\sum_{m=n}^\infty f(m)-f(n)$ for the inner sum
$$=\sum_{n=1}^\infty\left(\sum_{m=n}^\infty\frac{(-1)^{m-n}+(-1)^n}{nm^{k+3}}-\frac{1+(-1)^n}{n^{k+4}}\right)$$
$$=\sum_{n=1}^\infty\sum_{m=n}^\infty\frac{(-1)^{m-n}+(-1)^n}{nm^{k+3}}-\sum_{n=1}^\infty\frac{1+(-1)^n}{n^{k+4}}$$
use $\sum_{n=1}^\infty\sum_{m=n}^\infty f(n,m)=\sum_{m=1}^\infty\sum_{n=1}^m f(n,m)$ for the first term
$$=\sum_{m=1}^\infty\sum_{n=1}^m\frac{(-1)^{m-n}+(-1)^n}{nm^{k+3}}-\zeta(k+4)+\eta(k+4)$$
$$=\sum_{m=1}^\infty\frac{1}{m^{k+3}}\left(\sum_{n=1}^m\frac{(-1)^{m-n}+(-1)^n}{n}\right)-\zeta(k+4)+\eta(k+4)$$
$$=\sum_{m=1}^\infty\frac1{m^{k+3}}\left(-(-1)^m\overline{H}_m-\overline{H}_m\right)-\zeta(k+4)+\eta(k+4)$$
$$=-\sum_{m=1}^\infty\frac{(-1)^m\overline{H}_m}{m^{k+3}}-\sum_{m=1}^\infty\frac{\overline{H}_m}{m^{k+3}}-\zeta(k+4)+\eta(k+4)$$
By combining $S_1$ and $S_2$, the term $\sum_{m=1}^\infty\frac{(-1)^m\overline{H}_m}{m^{k+3}}$ nicely cancels out and we have
$$\sum_{j=0}^k\eta(k+2-j)\eta(j+2)$$
$$=(k+3)\zeta(k+4)-2\ln(2)[\eta(k+3)+\zeta(k+3)]
-2\eta(k+4)+2\sum_{m=1}^\infty\frac{\overline{H}_m}{m^{k+3}}$$
Letting $q=k+3$ and reindexing $j\mapsto j-1$ we arrive at
$$\sum_{j=1}^{q-2}\eta(q-j)\eta(j+1)
=q\zeta(q+1)-2\ln(2)[\eta(q)+\zeta(q)]-2\eta(q+1)+2\sum_{m=1}^\infty\frac{\overline{H}_m}{m^q}
$$
Write $\eta(s)=(1-2^{1-s})\zeta(s)$ we finally obtain
$$
\sum_{m=1}^\infty\frac{\overline{H}_m}{m^q}=\left(1-2^{-q}-\frac{q}{2}\right)\zeta(q+1)+(2-2^{1-q})\ln(2)\zeta(q)$$
$$+\frac12\sum_{j=1}^{q-2}(1-2^{1-g+j})(1-2^{-j})\zeta(q-j)\zeta(j+1)$$
Applications
\begin{equation}
\sum_{m=1}^\infty\frac{\overline{H}_m}{m^2}=\frac32\ln(2)\zeta(2)-\frac14\zeta(3)
\end{equation}
\begin{equation}
\sum_{m=1}^\infty\frac{\overline{H}_m}{m^3}=\frac74\ln(2)\zeta(3)-\frac5{16}\zeta(4)
\end{equation}
\begin{equation}
\sum_{m=1}^\infty\frac{\overline{H}_m}{m^4}=\frac{15}{8}\ln(2)\zeta(4)+\frac38\zeta(2)\zeta(3)-\frac{17}{16}\zeta(5)
\end{equation}
\begin{equation}
\sum_{m=1}^\infty\frac{\overline{H}_m}{m^5}=\frac{31}{16}\ln(2)\zeta(5)+\frac{9}{32}\zeta^2(3)-\frac{49}{64}\zeta(6)
\end{equation}
\begin{equation}
\sum_{m=1}^\infty\frac{\overline{H}_m}{m^6}=\frac{63}{32}\ln(2)\zeta(6)+\frac{21}{32}\zeta(3)\zeta(4)+\frac{15}{32}\zeta(2)\zeta(5)-\frac{129}{64}\zeta(7)
\end{equation}

Bonus:
By combining this generalization and $(1)$ from the question body and substituting the generalized euler sum we find
$$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^q}=q(1-2^{-1-q})\zeta(q+1)-(2-2^{1-q})\ln(2)\zeta(q)$$
$$-\frac12\sum_{j=1}^{q-2}(2^{j+1}-1)(2^{-j}-2^{-q})\zeta(q-j)\zeta(j+1)\label{H_n/(2n+1)^q}$$
Applications
\begin{equation}
\sum_{n=1}^\infty\frac{H_n}{(2n+1)^2}=\frac74\zeta(3)-\frac32\ln(2)\zeta(2)
\end{equation}
\begin{equation}
\sum_{n=1}^\infty\frac{H_n}{(2n+1)^3}=\frac{45}{32}\zeta(4)-\frac74\ln(2)\zeta(3)
\end{equation}
\begin{equation}
\sum_{n=1}^\infty\frac{H_n}{(2n+1)^4}=\frac{31}{8}\zeta(5)-\frac{15}{8}\ln(2)\zeta(4)-\frac{21}{16}\zeta(2)\zeta(3)
\end{equation}
\begin{equation}
\sum_{n=1}^\infty\frac{H_n}{(2n+1)^5}=\frac{315}{128}\zeta(6)-\frac{31}{16}\ln(2)\zeta(5)-\frac{49}{64}\zeta^2(3)
\end{equation}
\begin{equation}
\sum_{n=1}^\infty\frac{H_n}{(2n+1)^6}=\frac{381}{64}\zeta(7)-\frac{63}{32}\ln(2)\zeta(6)-\frac{93}{64}\zeta(2)\zeta(5)-\frac{105}{64}\zeta(3)\zeta(4)
\end{equation}